Why is $9<\sqrt{89}<10$? Explain why $9<\sqrt{89}<10$. How do you explain this? I'm doing revision and we haven't been taught it yet but it will be on the test.
$\sqrt{389}$ is also between two consecutive whole numbers.
What are the two numbers?
 A: Hint: what's $9^2$, and what's $10^2$?
A: I got it, it's $19^2=361$ and $20 =400$ for the two  consecutive numbers on both sides of $\sqrt{389}$.
A: Imagine a triangle (with one corner90) (with one corner( not 90 degree) in (0,0)point )  and one edge=$\sqrt8$ then other edge=9(on HORIZONTAL  axis ) then if you imagine a CIRCLE  with center (0,0)and $r=\sqrt{89}$ you will see that length of SINEW  of triangle ($\sqrt{89}$)will be larger than length of edge=9.
A: Hints:
$$9^2=81<89<100=10^2\ldots$$
Of course, you may also want to show the square root function $\,f(x):=\sqrt x\,$ is monotone ascending, for example by means of its first derivative....
A: Hint:
You want to demonstrate -
$9<\sqrt{89}<10$
What happens when you square across these terms?
This can be done because, based on the suggested edit by Cameron Buie, each term can be squared for an equivalent chain of inequalities since $x↦x^2$ is a strictly increasing function on the positive reals.
A: Recall that the square root is monotonically increasing function so
$$81<89<100\Rightarrow\sqrt{81}=9< \sqrt{89}<10=\sqrt{100}$$
A: In addition to the excellent answers above, I'd like to add this picture. (My tikz-fu needs all the practice I can give it).

A: $f(x)=x^2$ is a strictly increasing function on the positive real numbers, meaning that $$a<b\Leftrightarrow f(a)<f(b) \hspace{30pt} \forall \hspace{5pt} 0\leq a,b\in\mathbb{R}.$$
That means that it suffices to notice that $$\color{darkblue}{9^2=81}<\color{darkred}{89}<\color{purple}{100=10^2}$$ which therefore implies that $$\color{darkblue}{9}<\color{darkred}{\sqrt{89}}<\color{purple}{10}.$$
Here is some visual intuition on why the algebra works. The thick blue represents the graph of the function $f(x)=x^2$.

A: $$\begin{align*}
9<\sqrt{89}<10&\Rightarrow \sqrt{9^2}<\sqrt{89}<\sqrt{10^2}\\\,\\&\Rightarrow\sqrt{81}<\sqrt{89}<\sqrt{100}\,\,\,\checkmark
\end{align*}$$
